Count and Say Problem in DSA C - Time & Space Complexity

Time Complexity: Count and Say Problem

O(2^n)

Understanding Time Complexity

We want to understand how the time needed to generate the "count and say" sequence grows as we ask for more terms.

How does the work increase when we want the nth term?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

char* countAndSay(int n) {
    if (n == 1) return strdup("1");
    char* prev = countAndSay(n - 1);
    int len = strlen(prev);
    char* result = malloc(len * 2 + 1);
    int idx = 0, count = 1;
    for (int i = 1; i <= len; i++) {
        if (i < len && prev[i] == prev[i - 1]) {
            count++;
        } else {
            idx += sprintf(result + idx, "%d%c", count, prev[i - 1]);
            count = 1;
        }
    }
    free(prev);
    return result;
}
    

This code generates the nth term of the count and say sequence by building from the previous term.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursion calling countAndSay(n-1) and a loop that scans the previous string to build the current one.
• How many times: The recursion happens n times, and each loop runs over the length of the previous string.

How Execution Grows With Input

Each term is built by reading the previous term, which roughly doubles in length each time.

Input Size (n)	Approx. Operations
1	1 (base case)
5	About 16 operations (length grows roughly doubling each step)
10	About 512 operations (string length grows exponentially)

Pattern observation: The work grows exponentially because each term length roughly doubles, so the total work sums up exponentially.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed roughly doubles with each increase in n, growing very fast.

Common Mistake

[X] Wrong: "The time grows linearly with n because we just call the function n times."

[OK] Correct: Each call processes a string that grows exponentially in length, so the work per call increases a lot, not just the number of calls.

Interview Connect

Understanding how recursion and string growth affect time helps you explain and optimize similar problems in interviews confidently.

Self-Check

"What if we stored all previous terms to avoid repeated recursion? How would the time complexity change?"

Practice

1. What is the output of the first term in the Count and Say sequence?

easy

A. "1"

B. "11"

C. "21"

D. "1211"

Solution

1. Step 1: Understand the first term definition

   The first term in the Count and Say sequence is always "1" by definition.

2. Step 2: Confirm the output

   Since the first term is "1", the output is exactly "1".

3. Final Answer:

   "1" -> Option A

4. Quick Check:

   First term = "1" [OK]

Hint: First term is always "1" in Count and Say [OK]

Common Mistakes:

• Confusing first term with second term
• Thinking sequence starts with "11"
• Mixing up output format

2. Which of the following is the correct syntax to generate the next term from a given term char* term in C for the Count and Say problem?

easy

A. Use a loop to count consecutive characters and build a new string

B. Directly print the term without processing

C. Use recursion without counting characters

D. Sort the characters in the term before counting

Solution

1. Step 1: Understand the Count and Say logic

   The next term is generated by counting consecutive characters and describing them.

2. Step 2: Identify correct approach

   Using a loop to count consecutive characters and build a new string matches the problem's logic.

3. Final Answer:

   Use a loop to count consecutive characters and build a new string -> Option A

4. Quick Check:

   Loop + count consecutive chars = correct syntax [OK]

Hint: Count consecutive chars with loop to build next term [OK]

Common Mistakes:

• Skipping counting step
• Trying to sort characters
• Using recursion without counting

3. What is the output of the Count and Say sequence at term 4?

medium

A. "21"

B. "111221"

C. "1211"

D. "3112"

Solution

1. Step 1: Generate terms up to 4

   Term 1: "1"
   Term 2: "11" (one 1)
   Term 3: "21" (two 1s)
   Term 4: "1211" (one 2, two 1s)

2. Step 2: Confirm term 4 output

   The fourth term is "1211" describing term 3.

3. Final Answer:

   "1211" -> Option C

4. Quick Check:

   Term 4 = "1211" [OK]

Hint: Build terms stepwise: count then say [OK]

Common Mistakes:

• Skipping terms and guessing output
• Mixing term 3 and 4
• Incorrect counting of digits

4. Given the following C code snippet for Count and Say, what is the error?

char* nextTerm(char* term) {
    int count = 1;
    char result[100];
    int index = 0;
    for (int i = 0; term[i] != '\0'; i++) {
        if (term[i] == term[i+1]) {
            count++;
        } else {
            result[index++] = count + '0';
            result[index++] = term[i];
            count = 1;
        }
    }
    result[index] = '\0';
    return result;
}

medium

A. Loop condition should be i < strlen(term)

B. Incorrect character comparison operator

C. Missing initialization of count variable

D. Returning pointer to local array causes undefined behavior

Solution

1. Step 1: Analyze return of local array

   The function returns pointer to local array 'result' which is destroyed after function ends.

2. Step 2: Identify undefined behavior

   Returning local array pointer leads to undefined behavior as memory is invalid outside function.

3. Final Answer:

   Returning pointer to local array causes undefined behavior -> Option D

4. Quick Check:

   Local array pointer return = undefined behavior [OK]

Hint: Never return pointer to local array in C [OK]

Common Mistakes:

• Ignoring local array lifetime
• Assuming loop condition is wrong
• Overlooking pointer return issues

5. How would you modify the Count and Say algorithm to handle very large terms efficiently in C?

hard

A. Store all terms in a global array for quick access

B. Use dynamic memory allocation and process terms iteratively

C. Sort each term before counting to reduce complexity

D. Use fixed-size arrays and recursion for each term

Solution

1. Step 1: Understand large term challenges

   Large terms require flexible memory to avoid overflow and efficient processing.

2. Step 2: Choose suitable approach

   Dynamic memory allocation allows resizing and iterative processing avoids stack overflow.

3. Final Answer:

   Use dynamic memory allocation and process terms iteratively -> Option B

4. Quick Check:

   Dynamic memory + iteration = efficient large term handling [OK]

Hint: Use malloc and loops for big terms, avoid recursion [OK]

Common Mistakes:

• Using fixed arrays causing overflow
• Recursion causing stack overflow
• Sorting terms breaks sequence logic